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A useful motivating experiment moves a bead emitting fluorescence through the selected optical focus and records the detector signal against position. Measurement of point-spread function (PSF) for confocal fluorescence microscopy by InCheon Song and others (2005, §2 and Fig. 1) describes a $543.8\mathrm{nm}$ laser delivered through a single-mode optical fiber, collimation, beam-size adjustment and a dichroic beam splitter. A $100\times$, numerical aperture $1.3$ oil-immersion objective optical lens excites slide-mounted fluorescent beads and collects their emission. Emission passes a filter and collecting optical lens, then a pinhole, before an avalanche photodiode. A three-dimensional positioning stage moves the specimen. The apparatus also contains a phase plate for deliberately modified responses.^[ref]

The published comparison uses nominal bead diameters $0.17,0.5,1\mu \mathrm{m}$: larger beads give broader measured spots. This is a control on the finite emitter, because the image is the convolution of bead emission with the confocal point spread function. The smallest bead has weaker signal relative to noise; incorrect bead size, defocus, bleaching and optical aberrations limit inference from a restored response. The reported bead-size comparison concerns lateral responses; it is not a measured pinhole-size series or a measured sheet-sectioning curve.^[ref]

To test axial pinhole selection with this kind of apparatus, the following is a proposed extension, not an additional result from that paper. Choose an isolated bead small relative to the response, centre it laterally, then step $z$ through optical focus while recording background-subtracted signal at several pinhole settings. Hold excitation power, dwell interval, detector gain, emission band and sampling fixed. Preserve raw peaks before also plotting peak-normalized traces. A lower peak and faster relative decay with axial displacement would support the paired-selection mechanism; fitting only normalized widths would miss the throughput cost.

Measure blocked-excitation background, include a bead-free region, avoid detector saturation, and repeat scans in reversed order to reveal drift or bleaching. Check pinhole centring separately, since misalignment can mimic signal loss. Record bead diameter, immersion medium, refractive index, numerical aperture, wavelengths and axial step. A finite bead is an approximation to an optical point source; a point trace does not certify confocal optical sectioning of a whole fluorescent sheet. Depth-dependent spherical aberration, scattering and sample motion can broaden or distort the response. Increasing magnification or making a display brighter does not correct those effects.

Fourier's law predicts the local conductive heat flux from the local temperature gradient. In a continuum approximation with isotropic thermal conductivity $k>0$, its vector form and one-dimensional form areThe heat flux $q$ has units $\mathrm{W}{\mathrm{m}}^{-2}$, the temperature gradient has units $\mathrm{K}{\mathrm{m}}^{-1}$ and the thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. The rate of heat transfer $\dot{Q}$ has units $\mathrm{W}$. These units distinguish a local material response from the total power crossing a surface.

The minus sign makes heat flow toward lower temperature: for $k>0$, $\mathbf{q}\cdot \nabla T=-k|\nabla T{|}^2\le 0$. In the worked bar in thermal conduction, $dT/dx=-200\mathrm{K}{\mathrm{m}}^{-1}$ therefore gives $q=+400\mathrm{W}{\mathrm{m}}^{-2}$. Reversing the end temperatures reverses the heat flux. A zero temperature gradient gives zero conductive heat flux in this model.

Fourier's law is a constitutive relation expressing measured material behavior; it is not a consequence of conservation of energy alone. It applies locally and does not require a steady state. The straight-line bar solution additionally assumes constant thermal conductivity, constant cross-sectional area, no sources or side losses, and fixed end temperatures. Anisotropic materials require the directional response described by the thermal conductivity tensor. The scalar $k$ used in the examples expresses the stated isotropy assumption. ^[ref]

Historically, Joseph Fourier's Théorie analytique de la chaleur was published in 1822. In Chapter II, article 112, he writes the heat crossing a spherical surface as the product of thermal conductivity, elapsed time, surface area and the negative radial temperature gradient; article 113 balances the incoming and outgoing heat to obtain the heat equation for a sphere. The enduring idea is this connection between a local flow rule and a global prediction of changing temperature. ^{[ref] [ref]}

At a free termination of a taut string, zero transverse force imposes a Neumann boundary condition. Incident and reflected displacement amplitudes add with equal sign so their slopes cancel. The endpoint is a displacement antinode, all incident power is reflected, and the flux $-T{\psi }_x{\psi }_t$ through the boundary vanishes. This is a transverse-wave boundary condition, distinct from the quantum open-string usage of similar notation.

Consider a uniform plane wave of unit complex amplitude transmitted by an ideal optical slit of full width $D$. Treat the field as invariant along the optical slit height: this is one-dimensional transverse Fresnel propagation, so the prefactor is $1/\sqrt{i\lambda z}$, rather than the two-dimensional prefactor $1/(i\lambda z)$. Removing the carrier $e^{ikz}$ givesHere $I_{\mathrm{inc}}$ is incident irradiance. This infinitely long slit model describes a transverse cut away from the ends of a sufficiently long physical slit; its finite conserved quantity is transmitted power per unit optical slit height.

Put $t=\xi /D$ and $q=xD/(\lambda z)$. Substitution reduces the integral toThis form separates the pattern's shape from its physical scale. Reflection $t\mapsto -t$ proves $A(-q,N_D)=A(q,N_D)$, so the light intensity is an even function of $x$. At $N_D=0$, direct integration gives $A(q,0)=\sin (\pi q)/(\pi q)$, with limit one at $q=0$. Hence the Fraunhofer diffraction comparison is the squared sinc function. In this paragraph the unnormalized sinc function takes argument $\pi q$.

For an original example, choose wavelength $\lambda =633\mathrm{nm}$ and full width $D=0.400\mathrm{mm}$. At $z=0.0200\mathrm{m}$, $N_D\simeq 12.6382$, $N_a\simeq 3.15956$ and the edge-to-centre wave phase change is $9.92604$ radians. At $z=2.00\mathrm{m}$, $N_D\simeq 0.126382$, $N_a\simeq 0.0315956$ and that wave phase change is only $0.0992604$ radians. Even the nearer screen is fifty slit widths away; that geometric comparison alone misses its strong Fresnel diffraction.

These are numerical predictions from the displayed scalar integral, using the composite trapezoidal rule, rather than measured profiles. Each curve is divided by its own peak light intensity; the two horizontal scales differ. At 2 cm the computed central irradiance is about $1.14812I_{\mathrm{inc}}$, with brighter off-axis maxima. At 2 m it is about $0.126272I_{\mathrm{inc}}$ and is the peak. The dashed Fraunhofer diffraction curve has its own central normalization; its first zeros are at $x=\pm \lambda z/D=\pm 3.165\mathrm{mm}$. The finite-distance Fresnel diffraction curve remains slightly nonzero there.

Normalization hides the brightness change, but not the change in shape. The Fourier transform propagation multiplier has unit complex modulus, so Plancherel theorem givesThis is conservation of transmitted power per unit height in the ideal scalar model. An integral over only the displayed screen interval misses the tails; a peak-normalized profile is not a unit-integral probability density. Spreading and decreasing brightness together preserve the transmitted power.

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Frustrated total internal reflection occurs when a nearby second high-index dielectric couples to the evanescent field in total internal reflection, allowing power to cross a finite low-index gap. The extra boundary changes the dielectric-interface boundary conditions; the gap field need not be a single decaying evanescent wave. Consequently, the zero normal Poynting vector calculation for two infinite half-spaces does not forbid transmission in this three-medium geometry.

Harvard’s actual demonstration uses two right-angle glass prisms, a helium-neon laser, a small clamp and two perpendicular screens as detectors. The contact faces are cleaned and placed opposite each other, with the clamp protecting the corners. Increasing pressure narrows the intervening air film; the observed beam routing changes gradually from total internal reflection toward onward transmission. The broad-gap reflecting configuration supplies the comparison for the narrow-gap transmitting configuration. This is a qualitative readout: the page explicitly says the gap is not measured.^[ref]

The inference is that the lower-index region can mediate electromagnetic energy transfer when another boundary is nearby. The demonstration does not determine a numerical evanescent penetration depth, resolve gap uniformity, or establish a three-layer transmission coefficient. For the constructed $50^{\circ }$ example, $\delta \approx 178\mathrm{nm}$ identifies the scale over which the isolated-interface field remains appreciable; substituting $e^{-2g/\delta }$ for a gap transmission fraction would omit the changed boundary fields and their interference. A quantitative gap calculation must solve the full Maxwell equations with both boundaries.

A galvanic cell supplies electrical work from a spontaneous redox chemical reaction under the specified conditions. In discharge through a passive load, oxidation occurs at its negative anode and reduction at its positive cathode. Separating the reacting compartments makes the electrons take an external route, where their transport can deliver useful work. See Flowers et al., Chemistry 2e, §17.2: separated reactions and electrical work.

In the metallic external electrical circuit, electrons move from anode to cathode, while conventional current goes from cathode through the load to anode. Inside the cell, net ionic conventional current closes the loop from the anode compartment toward the cathode compartment. Oppositely moving cations and anions can both contribute to that direction; see electrolyte for the signed current density calculation. A cell can maintain a nonzero terminal electric potential difference near open circuit while delivering almost no power; the Daniell cell voltage and load experiment makes this distinction concrete.

Coupled dissolution of a less noble solid metal and reduction/deposition of a more noble metal ion without an external power supply.

For a narrow angular power pattern $\exp [-(u^2+w^2)/(2{\alpha }^2)]$ about an equatorial beam centre, the solid angle measure is locally $dudw$. Extending the small lobe to the plane gives its beam solid angle ${\Omega }_A\simeq 2\pi {\alpha }^2$. Both sides of the azimuthal lobe must be included. A lossless antenna then has peak antenna gain $G_0\simeq 2/{\alpha }^2$ and peak antenna effective area $A_0\simeq {\lambda }^2/(2\pi {\alpha }^2)$.

A fundamental Gaussian beam is an ideal beam of light whose transverse wave amplitude has a Gaussian function profile. Here consider a circularly symmetric beam in vacuum, with propagation along $z$, two transverse coordinates $x,y$, and radius $r=\sqrt{x^2+y^2}$ from the optical axis. With the beam waist at $z=0$, total power $P>0$, and vacuum wavelength $\lambda$, its time-averaged light intensity, meaning irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$, isHere $w_0$ is the minimum beam radius, and $z_R$ is the Rayleigh range. Far from the beam waist, $w(z)\sim \theta |z|$ with beam divergence half-angle $\theta =\lambda /(\pi w_0)$. These are the fundamental circular Gaussian beam laws within the scalar paraxial approximation, appropriate when $w_0\gg \lambda$ and the beam divergence is small. The underlying envelope solves the parabolic wave equation exactly; this does not make it an exact vector solution of Maxwell equations. ^[ref].

The radius convention matters. Writing $U$ for the complex wave amplitude envelope, the same $w(z)$ is the wave amplitude $1/e$ radius and the light intensity $1/e^2$ radius:At $r=w(z)$ these ratios are $e^{-1}\approx 0.367879$ and $e^{-2}\approx 0.135335$, respectively. The light intensity $1/e$ radius is instead $w(z)/\sqrt{2}$. None of these radii is a sharp edge: the ideal Gaussian function has nonzero tails at every finite radius. Kogelnik and Li explicitly use the wave amplitude $1/e$ radius in their fundamental-mode treatment. ^[ref].

For a numerical example, take vacuum wavelength $\lambda =632.8\mathrm{nm}$ and beam waist radius $w_0=0.500\mathrm{mm}$. Substitution gives Rayleigh range $z_R\approx 1.24114754\mathrm{m}$ and far-field beam divergence half-angle $\theta \approx 0.402852992\mathrm{mrad}$. At $z=z_R$, the beam radius is $w=\sqrt{2}{w}_0\approx 0.707107\mathrm{mm}$ and the on-axis irradiance is half its waist value. At the same wavelength, halving the beam waist radius to $0.250\mathrm{mm}$ quarters the Rayleigh range to $0.310286885\mathrm{m}$ and doubles the beam divergence half-angle to $0.805705984\mathrm{mrad}$. At fixed power, that smaller beam waist also quadruples the peak irradiance.

The contours in the left panel are $r=\pm w(z)$ in a plane through the optical axis; they mark an light intensity level, not an outer boundary. The right panel compares the exact model radii with the far-field asymptotes $w=\theta z$. The horizontal and vertical axes use different length units, so the drawn slopes are not physical angles.

The normalization expresses conserved power during free propagation. In transverse polar coordinates, an annulus has area $2\pi rdr$, soA proof outline is the substitution $s=2r^2/w(z{)}^2$ in this integral. Spreading reduces peak irradiance as $w^{-2}$ while increasing transverse area as $w^2$, leaving power unchanged. In particular, the disk of radius $w(z)$ contains $1-e^{-2}\approx 0.8646647168$, or $86.4665\%$, of the total power; the local light intensity at its rim is only $13.5335\%$ of the on-axis light intensity. These are different quantities.

The left panel compares normalized wave amplitude magnitude and irradiance; the right panel integrates the transverse area to show enclosed power. For an ideal absorbing circular aperture of radius $a$ centered on the optical axis, the same fraction gives the power passing the aperture at that plane. An aperture of radius $w$ clips about $13.5335\%$ of the incident power; radius $2w$ clips only $e^{-8}\approx 0.000335463$, or $0.0335463\%$. After clipping, extra diffraction means the transmitted field no longer obeys the unmodified ideal Gaussian beam law. Every finite aperture cuts some ideal tail. Kogelnik and Li distinguish their large-aperture beam analysis from their finite-aperture diffraction analysis. ^[ref].

The useful tradeoff is that a small bright spot cannot remain narrow indefinitely. An optical lens can create a smaller beam waist, but diffraction then makes the beam spread faster. The particularly simple relation $w_0\theta =\lambda /\pi$ ties localization directly to beam divergence. A proof outline for the propagation law starts with $U(x,y,0)=U_0{e}^{-(x^2+y^2)/w_0^2}$. The parabolic wave equation separates into two transverse factors, and a Gaussian integral in each coordinate givesHere $q$ is dimensionless and $k$ is the wavenumber; the carrier convention is $e^{i(kz-\omega t)}$, with angular frequency $\omega$. Taking the squared complex modulus gives the stated $w(z)$ and its $w^{-2}$ light intensity scaling. Each transverse factor contributes $q^{-1/2}$, explaining the $q^{-1}$ prefactor here and the different prefactor in the preserved Gaussian beam with one transverse coordinate.

Historically, Gaussian beam theory developed alongside early laser resonator theory and lens-guided transmission. In their 1966 review, Kogelnik and Li credit Boyd and Gordon, and Goubau and Schwering, with early beam treatments; they summarize earlier work rather than claim the whole construction as a new discovery. Their paper brings beam propagation, focusing and finite-aperture diffraction into a common practical account. ^[ref].

Choose an initially transform-limited pulse with Gaussian power and complex envelopeHere ${\sigma }_0$ is the standard deviation of the power-weighted time distribution; ${\tau }_0$ is its full width at half maximum. Neither is a field-amplitude width. The spectral amplitude is proportional to $e^{-{\sigma }_0^2{\Omega }^2}$, so the spectral power has standard deviation ${\sigma }_{\Omega }=1/(2{\sigma }_0)$. The Gaussian time-bandwidth product in ordinary frequency is ${\tau }_0\Delta \nu =2\ln 2/\pi \simeq 0.4413$, where both widths are power FWHMs.

Multiplication by $e^{i{\varphi }_2{\Omega }^2/2}$ preserves spectral power and energy. Substituting ${\varphi }^{\prime }={\varphi }_2\Omega$ into the moment identity givesThis is an exact result for a Gaussian spectrum with quadratic spectral phase. It also follows by completing the square in its inverse Fourier transform.

For these chosen inputs, ${\sigma }_0=12.7398\mathrm{fs}$. The stretched pulse has $\tau =278.88\mathrm{fs}$ and root mean square width $118.4282\mathrm{fs}$. Adding $-3000\mathrm{f}{\mathrm{s}}^2$ recovers $30\mathrm{fs}$; leaving $300\mathrm{f}{\mathrm{s}}^2$ gives $40.85\mathrm{fs}$. All phase-only cases retain $1\mathrm{nJ}$.

If GDD is cancelled but ${\varphi }_3=20000\mathrm{f}{\mathrm{s}}^3$ remains, the pulse is asymmetric. Gaussian spectral moments giveThus the centroid is $15.4033\mathrm{fs}$ and the root mean square duration is $25.2354\mathrm{fs}$. Converting this last value to a FWHM with the Gaussian factor would be wrong: the power envelope is no longer Gaussian. The calculation's integrated energy agrees in time and frequency through Parseval's identity, while its centroid and variance agree with these independent moment formulas.

A real amplifier also changes spectral amplitude. For a simple small-signal example, stipulate an intensity gain profileThis specifies finite gain bandwidth, with no saturation or extra amplifier phase. Multiplying two Gaussian spectra givesFor chosen ${\sigma }_g={\sigma }_{\Omega }$, the amplified spectrum is narrower by $\sqrt{2}$, the shortest Gaussian power FWHM becomes $30\sqrt{2}=42.43\mathrm{fs}$, and the total energy is $G_{0}U/\sqrt{2}$. Perfect spectral phase correction cannot recreate the suppressed spectral amplitude. Real gain narrowing and saturation require a material-dependent amplifier model; a constant multiplier is only a limited idealization. Gérard Mourou's 2018 lecture discusses gain narrowing as a limit on the shortest amplified pulses.^[ref]

For perturbation $V_0{e}^{-{\beta }^2{x}^2}$ and oscillator inverse length $\alpha =\sqrt{m\omega /\hslash }$, direct Gaussian expectation values give $\Delta E_0=V_0\alpha /({\alpha }^2+{\beta }^2{)}^{1/2}$ and $\Delta E_1=V_0{\alpha }^3/({\alpha }^2+{\beta }^2{)}^{3/2}$. A narrow perturbation samples the odd state's node and suppresses its shift by two extra powers of $\alpha /\beta$. A broad perturbation approaches a constant energy offset for both states.

Reversing the phase on one half of an otherwise uniform slit makes its aperture transmission odd. Its Fourier amplitude is $4{\sin }^2(qb/4)/(iq)$, where $q=k\sin \theta$, so the axis is dark despite the removable denominator singularity. The largest symmetric lobes satisfy $\tan s=2s$, with first positive root approximately $1.16556$. Doubling an ideal half-wave layer restores equal phases and the ordinary bright-centred single-slit pattern. Total transmitted power is conserved if the layer has no absorption or reflection.

A half-wave rectifier with one diode and a resistor load passes one half of each input cycle.^[ref]

For an original ideal example, let an ideal voltage source supply the sinusoid $v_S(t)=V_p\sin (\omega t)$, with $V_p=5\mathrm{V}$ and angular frequency $\omega >0$. Connect its positive reference terminal to the diode anode, the diode cathode to the upper terminal of a load resistor with electrical resistance $R_L=1\mathrm{k\Omega }$, and the lower terminal to the source return. Let $v_L$ be the upper-minus-lower load electric potential difference, and $i_L$ the downward load electric current. Assume the ideal diode model, zero source electrical resistance, and no filter capacitor, inductor or switching delay.

The Kirchhoff cycle law gives $v_S=V_D+v_L$, and Ohm's law gives $v_L=R_L{i}_L$. If $v_S>0$, the blocking assumption would give $V_D=v_S>0$, contradicting the ideal diode model, so the diode conducts with $V_D=0$. If $v_S<0$, conduction would give $i_L=v_S/R_L<0$, also forbidden, so the diode blocks. At the crossings both values vanish. ThusThe peak load electric current is $5\mathrm{mA}$. The output is unidirectional but still varies in time, reaching zero throughout every negative half-cycle. Over a period of a function $2\pi /\omega$, write the wave phase as $\theta =\omega t$. Its average value of a function and root mean square follow from two integrals:Consequently the average load power is $v_{L,\mathrm{rms}}^2/R_L=6.25\mathrm{mW}$, not ${\overline{v}}_L^2/R_L$. The ideal diode has zero loss; the resistor converts supplied energy into Joule heating. The same switching argument predicts $v_L=\min (0,v_S)$ when the diode is reversed. Directional selection changes the waveform and its average value of a function; it does not by itself produce a constant output.

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

A motivating bench experiment asks whether a hot bath receives more heat than the electrical work supplied to a heat pump. PHYWE's authored operating instructions, section 4, describe a sealed vapor-compression refrigeration apparatus with its cold and hot coils immersed in separate water vessels. The quantitative procedure calls for insulated vessels, stirring to reduce temperature gradients, timed temperature readings, and a wattmeter for the input. It waits about five minutes for operating conditions to settle before measurements and explicitly includes the glass vessel and immersed copper pipe when calculating the heated body's heat capacity. One bath cooling while the other warms makes the origin of the delivered energy observable. ^[ref]

The following analysis extends that procedure to simultaneous measurements of both finite baths. Weigh the water in each vessel; determine the effective heat capacities $C_c$ and $C_h$, including immersed hardware, by a separate known electrical-input calibration over the working temperature range. A matched pair of calibrated thermometers placed in stirred water measures bulk $T_c(t)$ and $T_h(t)$; keeping probes away from the coils avoids mistaking a local contact temperature for the whole bath. Record the room temperature and synchronise the electrical meter and thermometry. Use short intervals after startup, with no boiling or freezing and small enough temperature changes that constant effective heat capacities are a reasonable approximation. These are proposed measurement controls, rather than reported results from the manual.

Over a common interval $[t_0,t_1]$, the uncorrected bath estimates and measured electrical input areHere the wattmeter records real electrical power $P_{\mathrm{el}}$. The integral handles changing input, and the signs make both estimated useful heat quantities positive during heating and cooling. The effective heat capacity includes water mass times its specific heat capacity plus the response of the vessel and immersed hardware. Without that correction, equal temperature changes need not imply equal heat transfers.

Insulation reduces heat leaks but does not establish an isolated system. A warmer bath can lose heat to the room while a colder bath gains it. As a proposed control, measure pump-off drift at matched bath and room temperatures, with the same stirring and insulation. Writing $L_h>0$ for outward hot-bath leak power and $L_c>0$ for inward cold-bath leak power, the bath balances giveIntegrate these rates over the same interval before dividing by $W_{\mathrm{el}}$. For example, in a pump-off hot-bath control, $L_h=-C_{h}d{T}_h/dt$; in the cold-bath control, $L_c=C_{c}d{T}_c/dt$. Check that a stopped machine does not create a different leak path from the running one; otherwise use an independent bath calibration and keep that discrepancy as an uncertainty. Repeat at the same starting temperatures to distinguish drift from a change in COP as the baths separate.

In the illustrated toy model, $C_h=C_c=4500\mathrm{J}{\mathrm{K}}^{-1}$, $P_{\mathrm{el}}=50\mathrm{W}$, ${\dot{Q}}_h=150\mathrm{W}$, ${\dot{Q}}_c=100\mathrm{W}$, $L_h=15\mathrm{W}$, and $L_c=5\mathrm{W}$ are stipulated constants for $180\mathrm{s}$. The slopes are $d{T}_h/dt=0.030\mathrm{K}{\mathrm{s}}^{-1}$ and $d{T}_c/dt=-19/900\mathrm{K}{\mathrm{s}}^{-1}$. Consequently,The raw heating and cooling ratios are $2.7$ and $1.9$, whose difference is $0.8$. Correcting the bath leaks gives $3$ and $2$, whose difference is one. The apparent failure of $Q_h=Q_c+W$ was incomplete measurement of heat, not a violation of the first law of thermodynamics.

Real apparatus can also release motor heat and pipe heat directly to the room, and store internal energy during startup. For a machine boundary that receives electrical work, an additional outward heat transfer $Q_{\mathrm{room}}$ gives $Q_h+Q_{\mathrm{room}}=Q_c+W_{\mathrm{el}}-\Delta U_{\mathrm{machine}}$. The residual $Q_h-Q_c-W_{\mathrm{el}}$ therefore tests the boundary and corrections; do not force it to zero. Include any auxiliary electrical work in the input if quoting whole-apparatus COP, and account separately for stirring work deposited in a bath. Ignoring vessel heat capacity, sensor lag or room leaks can bias the inferred result. These limitations follow from extending the measured energy budget, and the manual's explicit hardware correction illustrates why such boundaries matter.

Finally, these vessels are finite bodies whose temperatures change; they are not fixed thermal reservoirs. Their ratio of integrated useful heat to integrated input work is an interval COP. A Carnot's theorem comparison using simultaneous bath temperatures is a local approximation only when the machine settles much faster than the baths change. An exact reversible comparison for changing bath temperatures instead uses the bath entropy changes, for example $C_h\ln (T_{h,1}/T_{h,0})+C_c\ln (T_{c,1}/T_{c,0})$ for constant heat capacities, negligible net machine entropy change and no room leaks. The experiment's beauty is that two thermometers, known heat capacities and an electrical-energy measurement expose both conservation of energy and the practical gap from reversibility.

Heat transfer asks how quickly heat crosses a boundary and how the temperature varies along its path. Its principal mechanisms are thermal conduction, convection and thermal radiation. A warm wall can conduct heat through its interior while exchanging heat with moving air and emitting thermal radiation at its surfaces. Identifying these paths lets us estimate heating requirements and cooling times.

Distinguish the accumulated heat $Q$ from its rate $\dot{Q}=dQ/dt$, a power measured in $\mathrm{W}$. The heat flux $q$ is the rate per unit area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$: a uniform normal heat flux across area $A$ gives $\dot{Q}=Aq$. The steady bar in thermal conduction carries $q=400\mathrm{W}{\mathrm{m}}^{-2}$ through $A=0.01{\mathrm{m}}^2$, hence $\dot{Q}=4\mathrm{W}$; maintained for $60\mathrm{s}$, it transfers $Q=240\mathrm{J}$. This distinction makes a surface's intensity and a whole object's heating load easy to compare.

A Hertzian dipole is an ideal electrically short oscillating electric dipole used to model radiation from a localized current element. In vacuum its far-field electric dipole radiation has ${\sin }^2\theta$ angular power distribution and total power $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$, evaluated at the retarded time. The electric-dipole approximation for radiation requires a source much smaller than its radiation wavelength.

A uniform electrically short current element of length $\ell$ with sinusoidal peak current $I_0$ has electric dipole moment amplitude $p_0=I_0\ell /\omega$. The mean electric dipole radiation power is ${\omega }^4{p}_0^2/(12\pi {ϵ}_0{c}^3)$, so fixing current makes it proportional to ${\ell }^2{\omega }^2$, rather than ${\ell }^2{\omega }^4$. Current, charge excursion and dipole moment are different quantities to hold fixed when comparing antennas. The approximation requires $\ell \ll \lambda$ and does not impose the nonuniform standing-current profile of a long antenna.

For a conducting disc of radius $a$ rotating in a uniform axial magnetic field $B$, the shaft-to-rim motional electromotive force is ${\int }_0^a\omega rBdr=\omega B{a}^2/2$. A current $I$ between shaft and rim gives the opposing torque $IB{a}^2/2$, making electromagnetic conversion power $\mathcal{E}I$ equal to mechanical torque times angular speed.